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You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

这道题总的来说不是很难。只需要历遍两个链表，将对应的值相加就可以了。

在计算的过程中要把进位计算在内，注意最后一位的进位。

下面直接贴代码：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode *result, *temp1, *temp2;        result = temp1 = NULL;        int carry = 0;  // 刚开始进位为0        while (l1 != NULL && l2 != NULL) {            temp2 = new ListNode((l1->val + l2->val + carry)%10);  // l1 + l2 的个位数            if (result == NULL) {                result = temp1 = temp2;            } else {                temp1->next = temp2;            }            temp1 = temp2;            carry = (l1->val + l2->val + carry)/10;            l1 = l1->next;            l2 = l2->next;        }        if (l1 != NULL) {            temp1->next = l1;        }        if (l2 != NULL) {            temp1->next = l2;        }        int i;        while (temp1-> next != NULL) {            temp1 = temp1->next;            i = temp1->val;            temp1->val = (i + carry)%10;            carry = (i + carry)/10;        }        if (carry != 0) {            temp2 = new ListNode(carry);            temp1->next = temp2;        }        return result;    }};

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